Question 767260
<pre>
1)
{{{x/(3*sqrt(2*x - 3)) - 1/sqrt(2*x - 3) = 1/3}}}
Multiplying both sides by {{{3*sqrt(2*x - 3)}}} we get
{{{x - 3 = sqrt(2*x - 3)}}}
Squaring both sides
{{{x^2 - 6*x + 9 = 2*x - 3}}}
or
{{{x^2 -8*x + 12 = 0}}}
Solving by factorization
{{{(x - 6)*(x - 2) = 0}}}
x = 6 or x = 2

2)
{{{(2*x^3 - 24)^(1/3) = -x}}}
Cubing both sides
{{{2*x^3 - 24 = (-x)^3 = -x^3}}}
Grouping like terms
{{{3*x^3 = 24}}}
{{{x^3 = 8}}}
{{{x = 2}}}

3)
{{{(4*x^4 - 48)^(1/4) = x}}}
Raising both sides to the power of 4
{{{4*x^4 - 48 = x^4}}}
Grouping like terms
{{{3*x^4 = 48}}}
{{{x^4 = 48/3 = 16}}}
Taking the 4th root
{{{x = 2}}} or {{{x = -2}}}

4)
{{{sqrt(x^2 - 8*x) = -3}}}
Squaring both sides
{{{x^2 - 8*x = 9}}}
{{{x^2 - 8*x - 9 = 0}}}
Solving by factorization
{{{(x - 9)*(x + 1) = 0}}}
x = 9 or x = -1


Hope you got it:)